Is sum of scaling operator invertible? Suppose you have $X = L^2(\mathbb{R})$ and for $f \in X$ and $0 < n < \infty$ (natural number) consider $g$ defined as
$$
g = \sum_{k=1}^{n} S_k f
$$
Where $S_k$ is defined as $(S_k f)(x) = f(kx)$ it is clear that $g \in X$. I was wondering if I defined the operator $T_n$ as
$$
T_n = \sum_{k=1}^{n} S_k
$$
Is such operator invertible? I'm almost tempted to say yes and I was trying to use Rudin's Corollary 2.12 b) in functional analysis we have

Corollary 2.12


a) If $\Lambda$ is a continuous linear mapping of an F-space $X$ onto an F-space $Y$,
then $\Lambda$ is open.


b) If $\Lambda$ satisfies (a) and is one-to-one, then $\Lambda^{-1} : Y \to X$ is continuous

However I'm not really sure about the one-to-one relationship and I cannot manage to find a counter example either. Can you help with a hint maybe?
Update:
I tried the following, but I still end up anywhere. If you pick a o.n. basis $\left\{ \varphi_j \right\}_{j \in \mathbb{Z}}$ And expand $g$ and $S_k f$ in this basis I end up with a set of equations
$$
a_j = \sum_{k=1}^n b_{j,k} \;\; j \in \mathbb{Z}
$$
So I suppose at least for surjection this would correspond to fix the sequence $\left\{ a_j \right\}$ and show that this can be obtained by the coefficient $\left\{ b_{j,k} \right\}$
As another attempt I've observed that if $F : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is the fourier operator then
$$
F T F^{-1} = \sum_{k=1}^n F S_k F^{-1} = \sum_{k=1}^n \frac{1}{k} S_{1/k}
$$
But I don't know if this helped either, it doesn't seem to me I got a simpler expression of the operator.
 A: We can show that $T_n$ is indeed one-to-one on $X$ for all $n$. The question of $T_n$ being onto seems harder to fully resolve, i.e. getting a definitive answer for all $n$. We can reduce it to a question about zeros of certain polynomials/finite Fourier series that, for individual $n$, is reasonably tractable computationally (not sure about analytically). And I think it's likely, based on the reduction, that $T_n$ is not onto for $n \geq 3$. But I'm not sure how to show this in general.

$T_n$ is one-to-one
Since $T_n$ acts separately on the portions of a function defined on the positive reals and the negative reals, it suffices to consider the action of $T_n$ on $L^2(\mathbb{R}^{+})$. And for this, the Mellin transform is a nice tool to use because of how it interacts with dilations. This transform is given by
$$\mathcal{M} f(s)
= \int_{0}^{\infty} x^{-1/2 + 2 \pi i s} f(x) \, dx,$$
and it's an isometry from $L^2(\mathbb{R}^{+})$ to $L^2(\mathbb{R})$. Note that for $\rho > 0$ we have
$$\mathcal{M}(S_{\rho} f)(s)
= \rho^{-1/2 - 2 \pi i s} \mathcal{M} f(s).$$
Based on this formula, we can write
$$\mathcal{M}(T_n f)(s)
= \mathcal{M} f(s) \sum_{k = 1}^{n} k^{-1/2 - 2 \pi i s}
= \mathcal{M} f(s) G_n(s).$$
So if $T_n f = 0$, we have $\mathcal{M} f \cdot G_n = 0$. From this it's not too hard to see that $T_n$ is one-to-one, since any zeros of $G_n$ are isolated.

Reducing the question of $T_n$ being onto
As to being onto, based on the calculation above we can see $T_n$ is onto if and only if $G_n$ is bounded away from $0$, i.e. if $1/G_n$ is bounded. That this is so for $n = 1, 2$ is pretty straightforward.
In the case that $n \geq 3$, we can reduce the problem in the following way. For primes $p \leq n$, let $z_p(s) = p^{-2 \pi i s} = e^{- 2 \pi i s \log p}$. Then we can write $G_n$ as a polynomial in the $z_p$. For example,
$$G_4(s)
= 1 + \frac{z_2(s)}{\sqrt{2}} + \frac{z_3(s)}{\sqrt{3}} + \frac{z_2(s)^2}{\sqrt{4}}.$$
We can then view $G_n$ as the composition of two maps: one from $\mathbb{R}$ to the torus $\mathbb{T}^{\pi(n)} \subseteq \mathbb{C}^{\pi(n)}$ and given by $s \mapsto (z_2(s), z_3(s), \ldots)$ (here $\pi(n)$ denotes the number of primes less than or equal to $n$); and the other from $\mathbb{C}^{\pi(n)}$ to $\mathbb{C}$, and given by the polynomial function as described above, call it $P_n$.
Now the image of the first map is dense in $\mathbb{T}^{\pi(n)}$, since the $\log p$ terms appearing in the exponents of the $z_p(s)$ are rationally independent. (This denseness follows from results for linear flow on the torus.) As a result, $G_n$ is bounded away from $0$ if and only if the second map, the polynomial one, has no zeros on the torus.

Zeros of $P_n$
It seems likely to me that $P_n$ has zeros on the torus for all $n \geq 3$, but I'm not sure how one would prove this in general. I did some checking and numerically confirmed there are zeros up to $n = 10$.
The $P_n$ can be viewed as finite Fourier series on the torus, so that's a possible connection to use. But a search in that vein didn't turn up anything that seemed especially relevant. There's also a definite number theoretic flavor to the $P_n$, but I don't have much knowledge in that area.
A: Let me try and flesh out my idea, as it is too long for a comment:
let us define a function $u(r)=f(e^r)$
Note that a scaling $(S_k f)(x) =f(kx)$ can be interpreted as a shift of $u$ for  $r=log(x); s_k=log(k)$,
namely $f(kx)=u(r+s_k)$.
Also note that $\sum_k u(r+s_k)=\int dt \sum_k\delta(t-s_k)u(r+t) dt$, that is,
a convolution $([\sum_k\delta(t-s_k)]*u(t))(r)$
Convolutions can be inverted via a Fourier transform, so if $\mathcal{F}[\sum_k\delta(t-s_k)]$ has zeros in the Fourier domain, it cannot be inverted. So the question becomes one of studying the zeros of $\mathcal{F}[\sum_k\delta(t-s_k)]$.
A: I believe the operator $T_{n}$ is injective. If you think about $f(kx)$ in relation to $f(x)$ then if $f(x)\geq 0$ for all $x\in\mathbb{R}$ then $f(kx)\geq 0$ for all $x\in\mathbb{R}$ and $k\in\mathbb{N}$. Likewise, if $f(x)\leq 0$ for all $x\in\mathbb{R}$ then $f(kx)\leq 0$ for all $x\in\mathbb{R}$ and for all $k\in\mathbb{N}$. So the only time you may have a function $f(x)$ with,
\begin{align}
\sum_{k=1}^{n}f(kx)=0
\end{align}
is if the $f(x)$ takes both negative and positive values. However, even in this case, the sum above would not be zero.
Suppose $I\subseteq\mathbb{R}$ is the support of $f(x)$. Suppose $(a,b)$ is a component of $I$ such that for all $x>b$, $f(x)=0$. That is, $(a,b)$ is the right most interval contained in $I$. Now consider $f(kx)$, then the component $(a,b)$ of $I$ becomes $(\frac{a}{k},\frac{b}{k})$. Clearly $\frac{b}{k}<b$, but if $\frac{b}{k}<a$ then $f(kx)=0$ in $(a,b)$ and so $f(x)+f(kx)\neq 0$ in $(a,b)$. Likewise, if $a<\frac{b}{k}<b$ then $f(x)+f(kx)\neq 0$ in $(\frac{b}{k},b)$. So there is always a set of positive measure such that,
\begin{align}
\sum_{k=1}^{n}f(kx)\neq 0
\end{align}
This is somewhat crude in the sense that I inherently assume that $(a,b)$ is in the positive half line. However, with a little more rigour you can consider the case where $(a,b)$ is a negative interval (in which case you take the left most interval contained in $I$) and in the special case where $I$ is an interval itself containing $0$ then you can split it as the union of two intervals and apply the previous argument to each subinterval.
The purpose of all this essentially shows that $T_{n}f=0\iff f=0$. Hence $\ker(T_{n})=\{0\}$ and so $T_{n}$ is injective.
The other question to ask, which you don't seem to have addressed in your post, is whether or not $T_{n}$ is surjective (I get the feeling it is not but I am open to someone showing me otherwise). In Rudin's corollary it seems he is assuming $\Lambda(x)$ is surjective, but this must still be shown.
EDIT: I just thought of the case where $f$ may be oscillatory with $f(x)\to 0$ as $|x|\to\infty$. In these cases I would think that taking an approximating sequence of complactly supported smooth functions for which we can apply the above argument and show the result holds for $f$ in the limit.
